User blog:HowStrongIs/Spring-Heeled Jack Gets a Makeover (The 3-D Battles of WorldRunner)
The 3d battles of worldrunner jack.gif Next up after 1943 is one of the original "3D" games, The 3-D Battle of WorldRunner. Aside from being a milestone for graphical abilities and Square Enix, it's also got some pretty good feats. Mainly because the game is entirely centered around running fast and jumping far. Plus, it's a pretty simple game to calc. There's only four feats in the whole game. The speed Jack runs at, the speed of Jack's jumps, the force of Jack's jumps, and the speed of Jack's blaster shots. All of these are fairly simple to calculate. Even simpler than the last few calculations. There's just one problem. Jack has no listed canon height. Buuuuuut, he does have a height we can scale. He appears in Chocobo Racing, which conveniently tells us how fast its racers are moving. By scaling the size of the finish line's squares based on how long it takes Jack to cross it we can then scale Jack's height from their width. At 49 mph, or 21.91 m/s, it takes Jack around 0.17 s to cross the finish line. This means that the finish line is 3.725 m long, which means each square of it is 1.863 m wide. In this image Jack is 88 px tall, while the block is 226 px wide. This makes Jack 0.7255 m tall. This doesn't measure his full height when standing up, of course, but that doesn't really matter. His posture is the same in WorldRunner. But anyways, we now have the one variable we really need to scale everything else here. Well, aside from his mass, which we'll assume to be around 20 kg, given the usual masses of people around this height and the fact he's wearing some kind of space suit. On to scaling his speed though. We can see in his normal game that Jack is still 88 px tall, while the blocks he runs over are 309 px wide. This makes each block, being a square, 2.547 m long. By going through the video frame by frame we can get the other variable needed for this equation, as we can see it takes Jack 0.19 s to cross each block. As for his jumps, all we need to scale is his jumping height and the time it takes him to jump in the air. Overlaying him at the beginning and peak of his arc shows that Jack jumps 296 px into the air. Comparing that to his height of 88 px shows that he's leaping 2.441 m into the air. Since Jack spends about 1.73 s in the air before landing that means he crossed around 23.19 m, since his forwards velocity is constant. However... the timeframe for his fall... doesn't actually match up with Earth's gravity. Which makes sense, since he's in space. But it's problematic, since we'll have to find out exactly what the gravity of this planet is. But we can get to that later, once the math actually starts. For now I just want to know how fast his energy shots are. The energy shots are actually really easy to determine, however. Just watching the video shows that it takes them about 0.07 s to cross a block. And we already know how long a block is, so. Yeah. Anyways though. It's time to do the math for this. ---- v = d/t *d = 2.547 m *t = 0.19 s 2.547/0.19 = 13.41 m/s This gives us his forwards velocity. How fast he goes when he's running. v = d/t *d = 2.441 m *t = 0.865 s 2.441/0.865 = 2.822 m/s This gives us his vertical velocity. How fast he goes up when he jumps. v = √((Vx)2+(Vy)2) *Vx = 13.41 m/s *Vy = 2.822 m/s √((13.41)2+(2.822)2) = 13.7 m/s And this tells us how fast he actually jumps. We need this for the force of his jumps, but in order to get that we first need the planet's gravity. a = (2d)/(t2) *d = 2.441 m *t = 0.865 s (2*2.441)/(0.8652) = 6.525 m/s^2 Under this gravity, his normal estimated mass of 20 kg would be bumped down to a weight of 130.5 N, or 13.3 kg as an equivalent. From here we just need the distance he accelerates over, which we'll assume to be 0.5 m. Now we just get his acceleration and, from there, the force he applies when jumping. a = v2-u2/2*s *v = 13.7 m/s *u = 0 m/s *s = 0.5 m 13.72-02/2*0.5 = 187.7 m/s^2 F = ma *m = 13.3 kg *a = 187.7 m/s^2 13.3*187.7 = 2496 N And lastly comes his blast speed, which is the easiest of all of these. v = d/t *d = 2.547 m *t = 0.07 s 2.547/0.07 = 36.39 m/s ---- Results *Spring-Heeled Jack Gets a Makeover (Running Speed) - 13.41 m/s *Spring-Heeled Jack Gets a Makeover (Jumping Speed) - 13.7 m/s *Spring-Heeled Jack Gets a Makeover (Jumping Force) - 2496 N or 254.5 kg-force *Spring-Heeled Jack Gets a Makeover (Blast Speed) - 36.39 m/s Category:Blog posts Category:NESlympics Category:Calculation